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BIOL/PBIO 3333
Quiz 3
11/4/13
Answers:
1. The fertile varieties are all multiples of 7: 14 (diploid), 28 (tetraploid), 42
(hexaploid), 56 (octaploid), etc.
Ans: b)
2. If the monoploid number is 7, 21 would be triploid. The chromosomes would
align along the metaphase plate and segregate randomly- one cell would get 1
homolog, one cell would get two, for each chromosome in the set (7).
Ans: c)
3. The disease is X-linked and being passed through the dad. The son must
therefore receive both the X and Y chromosome from the dad. They would
normally segregate from one another during 1st meiotic prophase; so
nondisjunction occurs in the father during the first division cycle
Ans: a)
4. Anhydrotic displasia shows mosaicism because of random X chromosome
inactivation during early embryogenesis- a result of Barr body formation and
dosage compensation in mammals. This is mediated by XIST gene expression
on the X undergoing inactivation- but it is the paternal mutant X activation that
leads to the mutant phenotype, so the maternal X is inactivated in mutant tissue.
Ans: e).
z+
x y
z
Pseudodominance:
phenotype of deletion
heterozygote is mutant
(-) for x and y; wild type
(+) for z
5.
g
1
2
3
4
5
6
7
r
n
b
a
v
m
x
c s
(n,r)
(a, m, v)
(g,r)
(c, m, v, x )
(b)
(c, s )
(m, x )
1) Start with the dele ons showing the smallest number of genes: e.g. in
chromosome 5, gene b must lie in a region that is missing in 5 but
present every else.
2) Use a similar logic to compare overlapping dele ons, star ng with the
ones containing the fewest genes: e.g. comparing chromosomes 1 and
3, the order must be g…r…n. Comparing chromosomes 4, 6 and 7, the
order must be c…s.
3) Using the genes already mapped, con nue the logic: e.g. comparing
chromosomes 2, 4, and 7, the order must be a…v…m….x.
6. (b): c and s are the closest genes to one another from this map.
7. Expected characteristics:
a) F+ will transfer chromosomal gene markers at LOW frequency (L = low
recombinants) only to F− cells
b) Hfr will transfer chromosomal gene markers at HIGH frequency (M = many
recombinants) only to F− cells
c) F− cells will yield no recombinants if crossed to themselves, but could show
either L or M colonies depending on whether they are crossed to F+ or Hfr
strains…
Using these criteria:
Strains 1,4 and 5 must be F- cells
Strain 2 and 8 must be Hfrs
Strains 3, 6, and 7 must be F+ cells
Ans: (a)
8. Strains 1 and 7 would represent a F- x F+ cross: The F- cells would be
converted to F+ cells.
Ans: (a) True
Agar
Type
1
2
3
4
Str S T U V
+
+
+
+
+
+
+
−
−
+
+
+
+
−
+
+
+
+
−
+
Agar type 1: selects for t+ recombinants
Agar type 2: selects for u+ recombinants
Agar type 3: selects for v+ recombinants
Agar type 4: selects for s+ recombinants
Timings Number of
of
Colonies of Agar
Samples of Type
1
2
3 4
t+ u+ v+ s+
0
2.5
5
7.5
10
12.5
15
17.5
20
25
30
35
0
0
0
10
60
110
140
165
180
182
185
187
0
4
60
132
220
315
370
398
414
416
420
425
0
0
0
0
0
0
0
0
0
4
35
40
0
0
0
0
0
6
67
104
125
138
140
142
9. Time of Entry:
u-t-s-v
Ans: e) none of the above
10.
Distance between s – t: 5 minutes
Distance between s – v: 12.5
minutes
Ans: (a) true
450
u
400
350
300
250
Series1
Series2
t
s
200
150
Series3
Series4
100
50
v
0
0
-50
5
10
15
20
25
30
35
40